How to graph magnitude plots for basic 4-pole filters?

I have developed magnitude plots for basic one-pole and two-pole filters derived from various Physics PDF's and tutorials I have found online and calculated from:

https://www.desmos.com/calculator/mjudrnexbo

But I cannot figure out what the correct plots for any simple four-pole resonant LPF, HPF, BP, & BR filters would be in the same terms.

Any help on what equivalent simple four equations would be? Thanks.

• there are a lot more degrees of freedom for 4-pole filters than for 2-pole filters. – robert bristow-johnson Dec 17 '18 at 6:38
• I'm looking for the most computationally simple 4-pole equations that will permit resonance for LPF/HPF and bandwidth for BP/BR to be controlled by a simple "Q" function. – user39558 Dec 18 '18 at 4:45
• Mike, each of the 2-pole filters that make up your 4-pole filter has their own independent Q. – robert bristow-johnson Dec 18 '18 at 5:31
• do you wanna do the Moog 4-pole filter? – robert bristow-johnson Dec 18 '18 at 5:31
• or do you want a 4-pole Butterworth or 4-pole Tchebyshev? – robert bristow-johnson Dec 18 '18 at 5:34

For Butterworth LP and HP

https://www.desmos.com/calculator/dxcnvc63av

Play the variable n.

There are several optimality criteria to choose from: Butterworth, Chebyshev (type $$1$$ and $$2$$), Cauer, etc., and each of them will give you a different magnitude response.

Butterworth filters have an especially simple magnitude (gain) function. For a low pass filter of order $$n$$ (i.e., with $$n$$ poles) you get

$$G_{LP}(f)=\frac{1}{\sqrt{1+\left(\frac{f}{f_c}\right)^{2n}}}\tag{1}$$

where $$f_c$$ is the $$3$$-dB cut-off frequency. The gain of an $$n^{th}$$ order Butterworth high pass filter is

$$G_{HP}(f)=\frac{\left|\frac{f}{f_c}\right|^{n}}{\sqrt{1+\left(\frac{f}{f_c}\right)^{2n}}}\tag{2}$$

The general formulas for band pass and band stop filters are more complicated. For a fourth order band pass filter you get

$$G_{BP}(f)=\frac{f_c^2f^2}{\sqrt{f^8-4f_0^2f^6+(f_c^4+6f_0^4)f^4-4f_0^6f^2+f_0^8}}\tag{3}$$

where $$f_0$$ is the center frequency:

$$f_0=\sqrt{f_{l}f_{u}}\tag{4}$$

with $$f_l$$ and $$f_u$$ the lower and upper cut-off frequencies, respectively. The frequency $$f_c$$ is given by

$$f_c=\frac{f_u^2-f_0^2}{f_u}\tag{5}$$

Finally, the magnitude of a fourth order Butterworth band stop filter is given by

$$G_{BS}(f)=\frac{(f_0^2-f^2)^2}{\sqrt{f^8-4f_0^2f^6+(f_c^4+6f_0^4)f^4-4f_0^6f^2+f_0^8}}\tag{6}$$

where the center frequency $$f_0$$ and $$f_c$$ are given by $$(4)$$ and $$(5)$$, respectively.

The figure below shows the magnitudes computed according to Eqs $$(1)$$, $$(2)$$, $$(3)$$, and $$(6)$$. The cut-off frequency for the low and high pass filters was chosen as $$f_c=4$$, and the lower and upper cut-off frequencies for the band pass and the band stop filters were chosen as $$f_l=2$$ and $$f_u=4$$, respectively, resulting in a center frequency $$f_0=2\sqrt{2}$$.

• Thank you very very much Matt. I need these equations to be resonant though with a Q value. Is there any chance you could provide the simplest equations that would allow a Q value? For bandpass/bandreject I'd ideally like the width of the band controlled by a simple Q value as well. Thanks so much if so. I do appreciate it. – user39558 Dec 18 '18 at 4:42
• Personally, I find it a bit simpler, in this case, to keep the lowpass prototype, and simply apply frequency transformations: $H(\frac x{f_p})$ for lowpass, $H(\frac{f_p}x)$ for HP, $H(\frac{x^2-f_c^2}{BW x})$ for BP, $H(\frac{BW x}{x^2-f_c^2})$ for BS. The only minor downgrade would be the need to precalculate $f_1$ and $f_2$ from solving the quadratics of BP and BS. – a concerned citizen Dec 18 '18 at 7:52
• @Mike: Fourth order systems generally don't have a simple $Q$ factor (like second order systems). What you can do is simply concatenate (multiply) two second-order systems with the same $Q$ factor to obtain a (very specific) fourth-order system. – Matt L. Dec 18 '18 at 11:30
• With LPF and HPF one could use this method ... dsprelated.com/showthread/comp.dsp/… – Juha P Dec 18 '18 at 15:21